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Math Help - length of a line

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    length of a line

    Find the length of the line of y = \frac {2}{3} x^\frac {3}{2} from x=0 to x=2

    i used the formula, i dont know if i was supposed to use u substitution, but i got 2 \sqrt{3}
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    Find the length of the line of y = \frac {2}{3} x^\frac {3}{2} from x=0 to x=2

    i used the formula, i dont know if i was supposed to use u substitution, but i got 2 \sqrt{3}
    sorry, that is incorrect ... how did you work it?
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    Quote Originally Posted by skeeter View Post
    sorry, that is incorrect ... how did you work it?
    i admire your help skeeter, i appreciate that you are helping me

    well i used the formula and got :

    \int \sqrt{1 + x^\frac {1}{2} dx } from 0 - 2 i am supposed to take the derivative of the initial equation and use it in the eq. yes?
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    i admire your help skeeter, i appreciate that you are helping me

    well i used the formula and got :

    \int \sqrt{1 + x^\frac {1}{2} dx } from 0 - 2 i am supposed to take the derivative of the initial equation and use it in the eq. yes?
    mistake in your "formula". here is the correct version ...

    S = \int_a^b \sqrt{1 + [f'(x)]^2 } \, dx

    the derivative is squared
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    Quote Originally Posted by skeeter View Post
    mistake in your "formula". here is the correct version ...

    S = \int_a^b \sqrt{1 + [f'(x)]^2 } \, dx

    the derivative is squared
    oh, i know, i squared it on my work here, but i still got the same answer...because i get 1 + x inside the radical, yes?
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    oh, i know, i squared it on my work here, but i still got the same answer...because i get 1 + x inside the radical, yes?
    what is your antiderivative and how did you evaluate it?
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    Quote Originally Posted by skeeter View Post
    what is your antiderivative and how did you evaluate it?
    i dont know what you mean, but from the eq.

    \int_0^2 \sqrt{1 + (x^\frac {1}{2})^2 dx}
    from there i got
    \int_0^2 \sqrt{1+x dx}

    then; \int_0^2 (1 + x)^ \frac{1}{2}

    and went on from there
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    i dont know what you mean, but from the eq.

    \int_0^2 \sqrt{1 + (x^\frac {1}{2})^2 dx}
    from there i got
    \int_0^2 \sqrt{1+x dx}

    then; \int_0^2 (1 + x)^ \frac{1}{2}

    and went on from there
    "went from there", how?
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    Quote Originally Posted by skeeter View Post
    "went from there", how?
    after that, i just took the antiderivative and got

     \frac {2}{3} (1 + x)^ \frac {3}{2} from 0 - 2
    i then substituted the values. is that ok?
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    after that, i just took the antiderivative and got

     \frac {2}{3} (1 + x)^ \frac {3}{2} from 0 - 2
    i then substituted the values. is that ok?
    keep going ... show me how you arrived at the value 2\sqrt{3}.
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    Quote Originally Posted by skeeter View Post
    keep going ... show me how you arrived at the value 2\sqrt{3}.
    then it comes up to be \frac {2}{3} (\sqrt{3})^3
    i end up with a fraction which is simplified to 2rad 3
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  12. #12
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    then it comes up to be \frac {2}{3} (\sqrt{3})^3
    i end up with a fraction which is simplified to 2rad 3
    ok ... \frac{2}{3}(1+2)^{\frac{3}{2}} = 2\sqrt{3}

    but ...

    what did you get when you evaluated \frac{2}{3}(1+x)^{\frac{3}{2}} using the lower limit of integration, x = 0 ?

    ... or did you just assume it would have a value of 0 ?
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  13. #13
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    Quote Originally Posted by skeeter View Post
    ok ... \frac{2}{3}(1+2)^{\frac{3}{2}} = 2\sqrt{3}

    but ...

    what did you get when you evaluated \frac{2}{3}(1+x)^{\frac{3}{2}} using the lower limit of integration, x = 0 ?

    ... or did you just assume it would have a value of 0 ?
    thats my mistake isnt it?

    now i get it...i worked to fast on the problem and disregarded the 1

    thank you
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